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M.C.Q (1 Marks)

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MCQ 11 Mark
[Co(NH3​)4​Cl2​] possesses:
  • A
    Square planar geometry
  • B
    Tetrahedral geometry
  • C
    Tetrahedral nature
  • D
    Octahedral geometry
Answer
  1. Octahedral geometry

Explanation:

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MCQ 21 Mark
Total number of stereoisomers of [Co(acac)2​BrCl]- are.
  • A
    4
  • B
    3
  • C
    6
  • D
    2
Answer
  1. 3

Explanation:

Total number of stereo isomers of [Co(acac)2​BrCl] are 3 i.e. cis, trans, optical.

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MCQ 31 Mark
The correct order of ligands for writing the formula of complex compounds is _________.
  • A
    Neutral, anionic, cationic
  • B
    Anionic, neutral, cationic
  • C
    Anionic, cationic, neutral
  • D
    Cationic, neutral, anionic
Answer
  1. Neutral, anionic, cationic

Explanation:

The formula of a coordination complex is written in a different order than its name. The chemical symbol of the metal center is written first.

The ligands are written next, with neutral ligands coming before anionic ligands. If there is more than one anion or neutral ligand, they are written in alphabetical order according to the first letter in their chemical formula.

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MCQ 41 Mark
The co-ordination number of a metal in co-ordination compound is:
  • A
    Same as primary valency.
  • B
    Sum of primary and secondary valences.
  • C
    Same as secondary valency.
  • D
    None of the above.
Answer
  1. Same as secondary valency.

Explanation:

The secondary valency is equal to the coordination number the secondary valency are non ionizable valencies. These are satisfied by neutral molecules or negative ions.

For example in [Ni(CO)4​] the coordination number of Ni metal is four and its secondary valency is also four.

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MCQ 51 Mark
IUPAC name of [Pt(NH3)2 Cl(NO2)] is:
  • A
    Platinum diaminechloronitrite.
  • B
    Chloronitrito-N-ammineplatinum (II).
  • C
    Diamminechloridonitrito-N-platinum (II).
  • D
    Diamminechloronitrito-N-platinate (II).
Answer
  1. Diamminechloridonitrito-N-platinum (II).

Explanation:

In this case both the central atom as well as ligands are present in the coordination sphere. The legands are named first in alphabetical order before the name of central atom or ion. Names of the anionic ligands end with suffix '-o'. The name of the central metal atom is written at the end along with its oxidation state in Roman numeral, which is (II) for the central metal atom "platinum". Note the "chlorido" is preferred term/ name over chloro for chloride ion as a ligand as per modern trend.

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MCQ 61 Mark
An ion M2+, forms the complexes [M(H2​O)6​]2+, [M(en)3​]2+ and [MBr6​]4−. The colour of the complexes will be _____________ respectively.
  • A
    Green, blue and red
  • B
    Blue, red and green
  • C
    Green, red and blue
  • D
    Red, blue and green
Answer
  1. Blue, red and green

Explanation:

Crystal Field Stabilization Energy(CFSE) is proportional to the frequency of the absorbed light. The emitted colors are red, green and blue. The corresponding absorbed colors are green, red and orange.

So the first complex will be blue, second is red and third is green.

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MCQ 71 Mark
IUPAC name for the complex [Cu(NH3​)4​]SO4​ is:
  • A
    Cupramonium sulphate.
  • B
    Copper sulphate tetraamonia.
  • C
    Tetrammine copper (II) sulphate.
  • D
    Copper amonium (IV) sulphate.
Answer
  1. Tetrammine copper (II) sulphate.

Explanation:

Since 4 amino groups are present, it will be called as tetrammine.

Since copper(II) is the central metal, it will be tetrammine copper(II). Note that we can determine the oxidation number of copper by charge balance.

Now finally since sulphate is the counter ion, it's full name will be tetrammine copper(II) sulphate.

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MCQ 81 Mark
Which compound is formed when excess of KCN is added to an aqueous solution of copper sulphate?
  • A
    Cu(CN)2​
  • B
    K2​[Cu(CN)2​]
  • C
    K[Cu(CN)2​]
  • D
    K3​[Cu(CN)4​]
Answer
  1. K3​[Cu(CN)4​]

Explanation:

When excess of KCN is added to an aqueous solution of CuSO4​ then-

Initially, cupric cyanide is formed, Cu(CN)2​.

CuSO4​ + 2KCN → K2​CO4 ​+ Cu(CN)2​

Cupric Cyanide will decompose to produce cuprous cyanide,Cu2​(CN)2​.

2Cu(CN)​→ 2Cu2​(CN)2​ + 2(CN)

Cuprous cyanide reacts with an excess of KCN to form a complex, K3​[Cu(CN)4​], which is a stable complex.

CuCN + 3Ka → K3​[Cu(CN)4​].

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MCQ 91 Mark
Which one of the following molecular geometries (i.e. shapes) is not possible for the sp3d2 hybridization?
  • A
    Capped octahedral
  • B
    Octahedral
  • C
    Square planar
  • D
    Square pyramidal
Answer
  1. Capped octahedral

Explanation:

 

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MCQ 101 Mark
The brown complex obtained in the detection of nitrate radical is formulated as [Fe(H2​O)5​NO]SO4​. What is the oxidation number of Fe in this complex?
  • A
    +1
  • B
    +2
  • C
    +3
  • D
    0
Answer
  1. +1

Explanation:

The brown colour complex obtained in the detection of nitrate radical is formulated as [Fe(H2​O)5​NO]SO4​. In this complex, oxidation state of iron is +1. The univalent character of iron in the complex is justified by the presence of a coordinated NO+ group.

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MCQ 111 Mark
Which of the following complexes is used to be as an anticancer agent?
  • A
    mer−[Co(NH3​)3​Cl]
  • B
    cis−[PtCl2​NH3​)2​]
  • C
    cis−K2​[PtCl2​Br2​]
  • D
    Na2​CoCl4
Answer
  1. cis−[PtCl2​NH3​)2​]

Explanation:

cis−[PtCl2​NH3​)2​] known as cis platin is used as an anticancer agent.

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MCQ 121 Mark
Linear combination of two hybridzed orbitals, belonging to two atoms and each having one electron leads to:
  • A
    Sigma - bond
  • B
    Double bond
  • C
    Co-ordinate covalent bond
  • D
    Pi - bond
Answer
  1. Sigma - bond

Explanation:

Linear combination of two hybridized orbitals, belonging to two atoms and each having one electron leads to formation of sigma bond.

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MCQ 131 Mark
Which of the following statement is correct?
  • A
    Energy difference should be more between orbitals which undergo hybridization.
  • B
    Number of hybrid orbitals formed should be same as the number of atomic orbitals involved in hybridization.
  • C
    Hybrid orbitals arrange around the center of the atom unsymmetrically.
  • D
    Hybrid orbitals can form π bonds.
Answer
  1. Number of hybrid orbitals formed should be same as the number of atomic orbitals involved in hybridization.

Explanation:

Number of hybrid orbitals formed is equal to number of atomic orbitals involved in hybridisation. $\pi$ bonds will be formed always by pure orbitals (p or d orbitals). Hybrid orbitals arrange around the centre of atom gymmetrically depending on number of bond pairs and lone pairs.

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MCQ 141 Mark
What is the correct name of IF7​?
  • A
    Iron heptafluoride
  • B
    Iodine heptafluoride
  • C
    Iodine septafluoride
  • D
    Iron (IV) septafluoride
Answer
  1. Iodine heptafluoride

Explanation:

Iodine heptafluoride, also known as iodine(VII) fluoride or iodine fluoride, is an interhalogen compound with the chemical formula IF7​. It has an unusual pentagonal bipyramidal structure.

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MCQ 151 Mark
Which of the following complexes formed by Cu2+ ions is most stable?
  • A
    $\text{Cu}^{2+} + 4\text{NH}_3\rightleftharpoons[\text{Cu}(\text{NH}_3)_4]^{2+},\ \log\text{K} = 11.6$
  • B
    $\text{Cu}^{2+} + 4\text{CN}^-\rightleftharpoons[\text{Cu}(\text{CN})_4]^{2-},\ \log\text{K} = 27.3$
  • C
    $\text{Cu}^{2+} + 2\text{en}\rightleftharpoons[\text{Cu}(\text{en})_2]^{2+},\ \log\text{K} = 15.4$
  • D
    $\text{Cu}^{2+} +4\text{H}_2\text{O}\rightleftharpoons[\text{Cu}(\text{H}_2\text{O})_4]^{2+},\ \log\text{K} = 8.9$
Answer
  1. $\text{Cu}^{2+} + 4\text{CN}^-\rightleftharpoons[\text{Cu}(\text{CN})_4]^{2-},\ \log\text{K} = 27.3$

Explanation:

The stability of complexes is represented by the magnitude of its "Stability constant (K)". A higher value of stability constant of certain complex denotes its greater stability. Mathematically, the same hols good for log K values. Among the given log K values of different complexes, it is noted that the log K value for complex at (ii) is maximum (ie = 27.3). Hence it is most stable.

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MCQ 161 Mark
TEL is a compound used as:
  • A
    Antibiotic
  • B
    Antiseptic
  • C
    Antiknocking
  • D
    None of these
Answer
  1. Antiknocking

Explanation:

Anti knocking compounds are the chemicals which reduce knocking for improving the quality of gasoline. example:- Tetraethyl lead (TEL).

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MCQ 171 Mark
Example for a coordination compound is ____________.
  • A
    KCl.MgCl2​.6H2​O
  • B
    K2​SO4​Al2​(SO3​)24H2​O
  • C
    CoCl3​.6NH3​
  • D
    FeSO4​.(NH4​)2​SO4​.6H2​O
Answer
  1. K2​SO4​Al2​(SO3​)24H2​O

Explanation:

Carnallite: KCl.MgCl2​.6H2​O
Alum: K2​SO4​Al2​(SO3​)24H2​O
Ferrous ammonium sulphate: (NH4​)2​Fe(SO4​)2​6H2​O.
All are double salts.
Only [Co(NH3​)6​].Cl3​ is complex compound.

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MCQ 181 Mark
Amongst the following, the most stable complex is:
    • A
      [Fe(H2O)6]3+
    • B
      [Fe(NH3)6]3+
    • C
      [Fe(C2O4)3]3–
    • D
      [FeCl6]3–
    Answer
    1. [Fe(C2O4)3]3–

    In each of the given complex, Fe is in +3 oxidation state. As C2O42-  is didentate chelating ligand, it forms chelate rings and hence (c) out of complexes given above is the most stable complex.

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    MCQ 191 Mark
    The formula of the complex, tris- (ethylenediamine) cobalt (iii) sulphate is:
    • A
      [Co(en)2​SO4​]
    • B
      [Co(en)3​SO4​]
    • C
      [Co(en)3​]2​(SO4​)3
    • D
      [Co(en)3​]2​SO4
    Answer
    1. [Co(en)3​]2​(SO4​)3

    Explanation:

    [Co(en)3​]2​(SO4​)3

    Oxidation of en = 0 SO= −2 then we get CO = +3 as given.

    And tris is used as two numbers are present at that same spot.

    where 2 is called "di" and 3 is called "tris".

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    MCQ 201 Mark
    Developed photography film is due to:
    • A
      AgBr
    • B
      [Ag(​S2O3​)2​]3+
    • C
      Ag
    • D
      Ag2​O
    Answer
    1. Ag

    Explanation:

    The developed photography film is due to Ag.

    The treatment of the exposed photgraphic film with reducing agent is called developing of the film.

    A developer is usually an alkaline solution of hydroquinone or alkaline solution of pyrogallol.

    The pairs activated by light are reduced to deposit more of black silver.

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    MCQ 211 Mark
    The value of X in Fe(CO)2​(NO)x, is:
    • A
      3
    • B
      4
    • C
      2
    • D
      1
    Answer
    1. 2

    Explanation:

    The coordination sphere of a coordination compound or complex consists of the central metal atom/ion plus its attached ligands. The coordination sphere is usually enclosed in brackets when written in a formula. The coordination number is the number of donor atoms bonded to the central metal atom/ion.

    Most metal complexes or compounds except for alloys. Specific examples include hemoglobin and Ru3​(CO)12​.

    So the total electron should be 18

    valence electron of iron is = 8

    CO gives = 2e, NO gives 3e

    18 = 8 + (2 × 2) + (3 × x)

    18 - 8 - 4 = 3x

    x = 2

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    MCQ 221 Mark
    The number of isomers exhibited by [Cr(NH3​)3​Cl3​] is:
    • A
      2
    • B
      3
    • C
      4
    • D
      5
    Answer
    1. 2

    Explanation:

    The complex [Cr(NH3​)3​Cl3​] is of the type Ma3​b3​, has two isomers cis and trans or facial and meridional.

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    MCQ 231 Mark
    The formula for lithium iodotris(trifluorophosphine)nickelate(0) is:
    • A
      Li[Ni(PF3​)3​]
    • B
      Li[Ni(PF3​)3​I3​]
    • C
      Li[(PF)3​I3​]
    • D
      None of these
    Answer
    1. None of these

    Explanation:

    Trifluorophosphine is PF3​.

    The formula should be Li[Ni(PF3​)3​I].

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    MCQ 241 Mark
    Which of the following complexes show linkage isomerism?
    • A
      [Co(NH3)5(NO2)]2+
    • B
      [Co(H2O)5CO]3+
    • C
      [Cr(NH3)5SCN]2+
    • D
      [Fe(en)2Cl2]+
    Answer
    1. [Co(NH3)5(NO2)]2+
    1. [Cr(NH3)5SCN]2+

    Explanation:

    NO2 and SCN are ambidentate ligands hence, show linkage isomerism.

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    MCQ 251 Mark
    Which of the following complexes are heteroleptic?
    • A
      [Cr(NH3)6]3+
    • B
      [Fe(NH3)4Cl2]+
    • C
      [Mn(CN)6]4-
    • D
      [Co(NH3)4Cl2]
    Answer
    1. [Fe(NH3)4Cl2]+
    1. [Co(NH3)4Cl2]

    Explanation:

    Complexes in which a metal is bound to more than one kind of donor groups are known as heteroleptic. This condition is satisfied by complexes mentioned at (ii) & (iv) where the donor groups are NH3 & Cl2. Therefore, complexes at (ii) & (iv) are heteroleptic.

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    MCQ 261 Mark
    Transition elements form complexes readily because:
    • A
      Small size of cation
    • B
      Large ionic Charge
    • C
      Vacant d orbitals
    • D
      All the above
    Answer
    1. All the above

    Explanation:

    Transition elements form complexes readily because

    Due to lower size and higher charge, they have high charge density.

    They have vacant d orbital in which a ligand can donate its electron and form complex.

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    MCQ 271 Mark
    The stability of ferric ion is due to:
    • A
      Half field f-orbital
    • B
      Half field d-orbital
    • C
      Complete field f-orbital
    • D
      Complete field d-orbital
    Answer
    1. Half field d-orbital

    Explanation:

    Fe3+ = 1s22s22p63s23p63d5

    d orbital is half filled.So ferric ion is stable.

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    MCQ 281 Mark
    The primary and secondary valency of a central metal ion in [Co(NH3​)4​CO3​]Cl complex is _________ and _________ respectively.
    • A
      2, 6
    • B
      4, 6
    • C
      4, 5
    • D
      3, 6
    Answer
    1. 3, 6

    Explanation:

    The primary and secondary valency of cobalt ion is 3 and 6 respectively.

    The charge on cobalt ion is +3 which balances −2 charge on carbonate ion and −1 charge on chloride ion. Hence, the primary valency of cobalt ion is 3.

    Cobalt ion is surrounded by four monodentate and one bidentate ligand (4 ammonia and on carbonate) in the coordination sphere. Hence, its coordination number (secondary valency) is 6.

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    MCQ 291 Mark
    Calculate the octahedral crystal field splitting energy in kJ/mol for [Fe(CN)6​]4−, if the wavelength of the most intensely absorbed light is 305nm.
    • A
      392kJ/mol
    • B
      643kJ/mol
    • C
      238kJ/mol
    • D
      456kJ/mol
    Answer
    1. 392kJ/mol
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    MCQ 301 Mark
    Hybridisation involves:
    • A
      Addition of an electron pair.
    • B
      Combination and redistribution of atomic orbitals.
    • C
      Removal of an electron pair.
    • D
      Separation of orbitals.
    Answer
    1. Combination and redistribution of atomic orbitals.

    Explanation:

    Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals.

    Atomic orbitals with similar energy combine to give degenerated hybrid orbitals.

    The number of combining atomic orbitals is equal to the hybridised orbitals.

    They involve redistribution of electrons of combined atomic orbital to hybrid orbitals.

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    MCQ 311 Mark
    Mg is an important component of which biomolecule occurring extensively in living world?
    • A
      Hemoglobin
    • B
      Chlorophyll
    • C
      Florigen
    • D
      ATP
    Answer
    1. Chlorophyll

    Explanation:

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    MCQ 321 Mark
    Barium ions, CN− and Co2+ form an ionic complex. If that complex is supposed to be 75% ionised in water with vant Hoff factor 'i' equal to four, then the coordination number of Co2+ in the complex can be.
    • A
      Six
    • B
      Five
    • C
      Four
    • D
      Six and Four both
    Answer
    1. Six

    Explanation:

    Complex formed is Bq2​[CO(CN)6​] in which oxidation of CO is (2+) and CN of CO is six.

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    MCQ 331 Mark
    Which of the following complex has five unpaired electrons?
    • A
      [Mn(H2​O)6​]2+
    • B
      [Mn(CN)6​]3−
    • C
      [CrCl3​(H2​O3​]
    • D
      [Ag(NH3​)2​]+
    Answer
    1. [Mn(H2​O)6​]2+

    Explanation:

    Mn2+ has d5 configuration, since H2​O is weak ligand field, so it does not causes pairing of electrons and hence it has 5 unpaired electrons.

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    MCQ 341 Mark
    Which of the following options are correct for [Fe(CN)6]3- complex?
    • A
      d2sp3 hybridisation.
    • B
      sp3d2 hybridisation.
    • C
      Paramagnetic.
    • D
      Diamagnetic.
    Answer
    1. d2sp3 hybridisation.
    1. Paramagnetic.

    Explanation:

    [Fe(CN)6]3-

    Magnetic nature-paramagnetic.

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    MCQ 351 Mark
    Which of the following pairs of complexes whose aqueous solutions gives pale yellow and white precipitates respectively with 0.1M AgNO3​?
    • A
      [Pt(NH3​)4​Br2​]Cl2​ and [Pt(NH3​​)4​Cl2]Br2
    • B
      [Co(NH3​)5​NO3​]Br and [Co(NH3​)5​Br]NO3​
    • C
      [Pt(NH3​​)4​Cl2]Br2​ and [Pt(NH3​)4​​Br2​]Cl2
    • D
      [Co(NH3​​)5​NO3​​]Cl and [Co(NH3​​)5​Cl]NO3​
    Answer
    1. [Pt(NH3​​)4​Cl2]Br2​ and [Pt(NH3​)4​​Br2​]Cl2

    Explanation:

    The complex having Br and Cl out of coordination sphere only gives AgBr(pale yellow ) and AgCl (white ppt.) respectively upon reaction with AgNO3​.

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    MCQ 361 Mark
    What kind of isomerism exists between [Cr(H2O)6]Cl3 (violet) and [Cr(H2O)5Cl]Cl2⋅H2O (greyish-green)?
    • A
      Linkage isomerism.
    • B
      Solvate isomerism.
    • C
      Ionisation isomerism.
    • D
      Coordination isomerism.
    Answer
    1. Solvate isomerism.

    Explanation:

    The given compounds havfe different number of water molecules inside and outside the coordinate sphere.

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    MCQ 371 Mark
    Which can exist both as diastereoisomer and enantiomer?
    • A
      [Pt(en)3​]4+
    • B
      [Pt(en)2​ClBr]2+
    • C
      [Ru(NH3​)4​Cl2​]0
    • D
      [PtCl2​Br2​]0
    Answer
    1. [Pt(en)3​]4+

    Explanation:

    It exists as lis trans which is diastereomer and its cis form is optically active and exits as enantiomer also, where as only exists as enantiomers.

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    MCQ 381 Mark
    1 mole of amino cobalt chloride on treating with an excess of AgNO3​ solution gives 2 moles of AgCl precipitate. The number of Cl ions which satisfies both primary and secondary valency of cobalt in the complex is:
    • A
      2
    • B
      3
    • C
      1
    • D
      0
    Answer
    1.  1

    Explanation:

    1 mole of amino cobalt chloride gives 2 moles of AgCl so number of Cl ions which satisfies both primary and secondary valency of cobalt in the complex is 1.

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    MCQ 391 Mark
    The formula of prussian blue is:
    • A
      Na4​[Fe(CN)6​]
    • B
      Fe(CN)2​
    • C
      Fe(CNS)3​
    • D
      Fe4​[Fe(CN)6​]3
    Answer
    1. Fe4​[Fe(CN)6​]3

    Explanation:

    Ferric ferrocyanide / Potassium ferrocyanide / Sodium ferrocyanide is prussian blue. 

    Fe4​[Fe(CN)6​]3

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    MCQ 401 Mark
    Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:
    • A
      K[Cr(H2​O)2​(C2​O4​)2​]
    • B
      [Co(en)3​]Cl3​
    • C
      [Co(NH3​)5​(NO2​)](NO3​)2​
    • D
      [Pt(NH3​)(H2​O)Cl2​]
    Answer
    1. K[Cr(H2​O)2​(C2​O4​)2​]

    Explanation:

    Exhibits cis trans.

    Exhibits optical Isomerism.

    Exhibits linkage Isomerism (NO2​,−ONO)

    Cis-Trans Isomerism.

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    MCQ 411 Mark
    Which of the following inert gas has the smallest ionization energy?
    • A
      He
    • B
      Ne
    • C
      Ar
    • D
      Xe
    Answer
    1. Xe

    Explanation:

    Ionization energy decreases from top to bottom of each column because the outer shell electrons become farther removed from the nucleus.

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    MCQ 421 Mark
    The correct IUPAC name of [Pt(NH3)2Cl2] is:
    • A
      Diamminedichloridoplatinum (II).
    • B
      Diamminedichloridoplatinum (IV).
    • C
      Diamminedichloridoplatinum (0).
    • D
      Dichloridodiammineplatinum (IV).
    Answer
    1. Diamminedichloridoplatinum (II).

    Explanation:

    The complex compound is [Pt(NH3)2Cl2]. NH3 is a neutral ligand and named as amine. Cl is anion ligand and named as chloride. Di or tri is prefixed to represent the number. So the name of the compound is- Diamminedichloridoplatinum (II).

    (II) represent the oxidation state of platinum.

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    MCQ 431 Mark
    Among the following coordination compounds, the one giving a white precipitate with BaCl2​ solution is:
    • A
      [Cr(H2​O)5​Br]SO4​
    • B
      [Cr(H2​O)5​SCN]
    • C
      [Co(NH3​)5​SO4​]Br
    • D
      [Pt(NH3​)6​]Cl4
    Answer
    1. [Cr(H2​O)5​Br]SO4​

    Explanation:

    Among the following coordination compounds, the one giving a white precipitate with BaCl2​ solution is [Cr(H2​O)5​Br]SO4​.

    It dissociates in aqueous solution to give sulphate ions. Sulphate ions react with barium chloride to form a white precipitate of barium sulphate.

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    MCQ 441 Mark
    The primary and secondary valency of Co in the octahedral complex CoCl3​.5NH3​ are __________.
    • A
      6 and 3
    • B
      3 and 6
    • C
      4 and 3
    • D
      3 and 5
    Answer
    1. 3 and 6

    Explanation:

    The primary valency of Co is 3 and the secondary valency of Co is 6.

    The primary valency represents the number of ions in the ionization sphere and the secondary valency represents the number of ligands.

    In this example, one chlorine atom serves dual role.

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    MCQ 451 Mark
    Number of Cl ions satisifying both primary and secondary valency are in CoCl3​.5NH3​.
    • A
      1
    • B
      2
    • C
      3
    • D
      4
    Answer
    1. 1

    Explanation:

    There is one Cl ion which satisfies both primary and secondary valency in CoCl3​.5NH3​.

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    MCQ 461 Mark
    The number of isomers for the compound with the molecular formula C2​BrClFI is:
    • A
      3
    • B
      4
    • C
      5
    • D
      6
    Answer
    1. 4

    Explanation:

    Total 6 isomers are possible.

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    MCQ 471 Mark
    Identify the optically active compounds from the following:
    • A
      [Co(en)3]3+
    • B
      Trans– [Co(en)2Cl2]+
    • C
      Cis– [Co(en)2Cl2]+
    • D
      [Cr(NH3)5Cl]
    Answer
    1. [Co(en)3]3+
    1. Cis– [Co(en)2Cl2]+

    Explanation:

    (a, c) [Co(en)3]3+ and cis [Co(en)2Cl2]2+ are optically active compounds because their mirror images are non-superimposable isomer.

    Non-supermposable isomers of [Co(en)3]3+

    Non-superimosable isomers of [Co(en)2Cl2]+

    Hence, (a) and (c) are correct choices.

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    MCQ 481 Mark
    The two compounds pentaamminesulphatocobalt (III) bromide and pentaamminesulphatocobalt (III) chloride represent:
    • A
      Linkage isomerism
    • B
      Ionization isomerism
    • C
      Coordination isomerism
    • D
      No isomerism
    Answer
    1. No isomerism

    Explanation:

    Pentaamminesulphatocobalt (III) bromide is [Co(SO4​)(NH3​)4​]Br and pentaamminesulphatocobalt (III) chloride is [Co(SO4​)(NH3​)5​]Cl.

    The two compounds pentaamminesulphatocobalt (III) bromide and pentaamminesulphatocobalt (III) chloride represent no isomerism.

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    MCQ 491 Mark
    Coordination compounds have great importance in biological systems. In this context which of the following statements is incorrect?
    • A
      Carboxypeptidase-A is an enzyme and contains zin.
    • B
      Haemoglobin is the red pigment of blood and contains iron.
    • C
      Cyanocobalamin is B12 and contains cobalt.
    • D
      Chlorophylls are green pigments in plants and contain calcium.
    Answer
    1. Chlorophylls are green pigments in plants and contain calcium.

    Explanation:

    Chlorophylls are green pigments in plants and contain magnesium.

    They are useful during photosynthesis to store energy in the form of glucose from carbon dioxide and water in presence of sunlight.

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    MCQ 501 Mark
    When KCN is added to CuSO4​ solution, there is formation of the stable water soluble complex. This complex is:
    • A
      K4​[Cu(CN)6​]
    • B
      K3​[Cu(CN)4​]
    • C
      K2​[Cu(CN)4​]
    • D
      K[Cu(CN)3​]
    Answer
    1. K3​[Cu(CN)4​]

    Explanation:

    When KCN is added to CuSO4​ solution, the following reaction occurs.

    4KCN+CuSO4​ → K3​[Cu(CN)4​].

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    M.C.Q (1 Marks) - Chemistry STD 12 Science Questions - Vidyadip